50ml of 0.1M NH4OH ,25ml NH₄Cl weere used to make a buffer . what is the PH is if Pk_bis 4.8?

The given buffer is a base buffer

[NH₄OH] =[(50*0.1)/75]

[NH₄Cl] =[(25*2)/75]

From the principle V₁M₁=V₂M₂

For a base buffer p^OH = p^Kb + log [Salt/Base]

Therefore p^OH 4.8+ log [{(25*2)/75}/{(50*0.1)/75}] = 4.8+ log (10) = 4.8+1 =5.8

Therefore pH = (14-pOH) = (14-5.8) = 8.2